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### Network # Linear Power supply

D

#### David Griffith

Jan 1, 1970
0
Hi all,

I am trying to construct a variable power supply from the schematic
listed here.

http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm#317.gif

Basically I wanted to know if the author has got the value for the
filter capacitor correct or not, as it seems very high (36 millifarads)

I have tried plugging in the maths and it gives me a more realistic
value, but I would appreciate any insight into which of us has got it
wrong...

"Electronic Devices (2nd Ed)" by Thomas L Floyd gives the equation for a
filter capacitors size to be roughly as follows

C = 0.0024 / (R * r)

where R is the load resistance and r is the 'ripple factor' or ripple
voltage over DC voltage.

The transformer I am using is not marked, when connected to a supply
measuring 240V rms exactly on my DVM measures 13.0V rms output on the
secondary.

I therefore take the peak voltage to be 13 * sqrt(2) = 18.4V peak

The LM317T regulator seems to require r to be 1/18.4 or less giving a
maximum actual ripple of 1V , and can output up to 1.5A setting R to be
V/I = 18.4/1.5 = 12.3 ohms

C = (0.0024 * 1.5) after cancelling voltage from both sides of the
equation = 3600uF not 36,000uF as stated in the schematics.

Is my maths sound?

David Griffith

J

#### John Popelish

Jan 1, 1970
0
David said:
Hi all,

I am trying to construct a variable power supply from the schematic
listed here.

http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm#317.gif

Basically I wanted to know if the author has got the value for the
filter capacitor correct or not, as it seems very high (36 millifarads)

I have tried plugging in the maths and it gives me a more realistic
value, but I would appreciate any insight into which of us has got it
wrong...

"Electronic Devices (2nd Ed)" by Thomas L Floyd gives the equation for a
filter capacitors size to be roughly as follows

C = 0.0024 / (R * r)

where R is the load resistance and r is the 'ripple factor' or ripple
voltage over DC voltage.

The transformer I am using is not marked, when connected to a supply
measuring 240V rms exactly on my DVM measures 13.0V rms output on the
secondary.

I therefore take the peak voltage to be 13 * sqrt(2) = 18.4V peak

The LM317T regulator seems to require r to be 1/18.4 or less giving a
maximum actual ripple of 1V , and can output up to 1.5A setting R to be
V/I = 18.4/1.5 = 12.3 ohms

C = (0.0024 * 1.5) after cancelling voltage from both sides of the
equation = 3600uF not 36,000uF as stated in the schematics.

Is my maths sound?

David Griffith

The supply with the 36,000 uf cap is a 4.3 amp version that needs lots
of voltage reserved for the pass transistor (5.5 volts at full load).
That big cap is there to keep the ripple from falling below the
required minimum. It is not a very good design, but that is what it
needs.

C

#### cpemma

Jan 1, 1970
0
David said:
Hi all,

I am trying to construct a variable power supply from the schematic
listed here.

http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm#317.gif

Basically I wanted to know if the author has got the value for the
filter capacitor correct or not, as it seems very high (36

I have tried plugging in the maths and it gives me a more realistic
value, but I would appreciate any insight into which of us has got it
wrong...

"Electronic Devices (2nd Ed)" by Thomas L Floyd gives the equation
for a filter capacitors size to be roughly as follows

C = 0.0024 / (R * r)

where R is the load resistance and r is the 'ripple factor' or ripple
voltage over DC voltage.

The transformer I am using is not marked, when connected to a supply
measuring 240V rms exactly on my DVM measures 13.0V rms output on the
secondary.

I therefore take the peak voltage to be 13 * sqrt(2) = 18.4V peak

The LM317T regulator seems to require r to be 1/18.4 or less giving a
maximum actual ripple of 1V , and can output up to 1.5A setting R to
be V/I = 18.4/1.5 = 12.3 ohms

C = (0.0024 * 1.5) after cancelling voltage from both sides of the
equation = 3600uF not 36,000uF as stated in the schematics.

Is my maths sound?

Using the simpler formula C = load current/(frequency x ripple voltage),
with 1.5A, our full-wave 100Hz and 1V pp ripple, gives 15,000uF, and with
Bill's 4.3A and US 120Hz it comes out at 35,833uF.

Formula from the guide at http://my.integritynet.com.au/purdic/power1.html

R

#### Rich Grise

Jan 1, 1970
0
David Griffith said:
Hi all,

I am trying to construct a variable power supply from the schematic
listed here.

http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm#317.gif

Basically I wanted to know if the author has got the value for the
filter capacitor correct or not, as it seems very high (36 millifarads)

Probably, since he says:
"To keep the ripple voltage below 1 volt at 4.3 amps, a 36,000 uF or greater
filter capacitor is needed."

Have Fun!
Rich

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